Percentage - Aptitude MCQ Questions and Solutions with Explanations

Let the weight of the bucket when it is full, be 1 kg.
Then, weight of empty bucket
= 25% of 1 kg
= $$\frac{1}{4}$$ kg
Weight of liquid in the bucket
= 1 - $$\frac{1}{4}$$ kg
= $$\frac{3}{4}$$ kg
On removing the liquid, weight of (bucket + liquid) = $$\frac{3}{5}$$ kg
Weight of liquid in the bucket
= $$\frac{3}{5}$$ - $$\frac{1}{4}$$ kg
= $$\frac{7}{20}$$ kg
Weight of liquid removed
= $$\frac{3}{4}$$ - $$\frac{7}{20}$$
= $$\frac{8}{20}$$
= $$\frac{2}{5}$$ kg
Hence, required percentage
= $$\frac{2}{5}$$ × $$\frac{4}{3}$$ × 100%
= $$\frac{160}{3}$$%
= $$53\frac{1}{3}$$%

Let total population = x
Then, number of males = $$\frac{5}{9}$$x
Married males :
= 30% of $$\frac{5}{9}$$x
= $$\frac{30}{100}$$ × $$\frac{5}{9}$$x
= $$\frac{x}{6}$$
Married females = $$\frac{x}{6}$$;
Number of females :
= x - $$\frac{5}{9}$$x
= $$\frac{4x}{9}$$
Unmarried females :
= $$\frac{4x}{9}$$ - $$\frac{x}{6}$$
= $$\frac{5x}{18}$$
∴ Required percentage :
= $$\frac{5x}{18}$$ × $$\frac{1}{x}$$ × 100%
= $$27\frac{7}{9}$$%

5 : 4
= $$\frac{5}{4}$$
= $$\frac{5}{4}$$ × 100%
= 125%

Given (550% of 250) ÷ 275 = (?)
⇒ (?) = $$\frac{550 × 250}{100}$$   ÷ 275
⇒ (?) = (55 × 25) ÷ 275
⇒ (?) = $$\frac{55 × 25}{275}$$
⇒ (?) = 5

Pass percentage :
$$\eqalign{ & = \left( {\frac{{252}}{{270}} \times 100} \right)\% \cr & = \frac{{280}}{3}\% \cr & = 93\frac{1}{3}\% \cr} $$

x% of y = $$\frac{xy}{100}$$..... (i)
And y% of x = $$\frac{xy}{100}$$ ..... (ii)
From (i) and (ii)
x% of y = y% of x

Let X and Y denote the new values of x and y respectively.
Then, X = 80% of x = $$\frac{4x}{5}$$
Y = 80% of y = $$\frac{4y}{5}$$
$$\eqalign{ & \therefore X{Y^2} = \frac{{4x}}{5} \times {\left( {\frac{{4y}}{5}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4x}}{5} \times \frac{{16{y^2}}}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{64}}{{125}}x{y^2} \cr} $$
Decrease in the value :
$$\eqalign{ & = \left( {x{y^2} - \frac{{64}}{{125}}x{y^2}} \right) \cr & = \frac{{61}}{{125}}x{y^2} \cr} $$
∴ Decrease %
$$\eqalign{ & = \frac{{61}}{{125}}x{y^2} \cr & = \left( {\frac{{61x{y^2}}}{{125}} \times \frac{1}{{x{y^2}}} \times 100} \right)\% \cr & = 48.8\% \cr} $$

Purchase price :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left[ {\frac{{29644.032}}{{{{\left( {1 - \frac{{12}}{{100}}} \right)}^3}}}} \right] \cr & = {\text{Rs}}{\text{.}}\left( {29644.032 \times \frac{{25}}{{22}} \times \frac{{25}}{{22}} \times \frac{{25}}{{22}}} \right) \cr & = {\text{Rs}}.43500 \cr} $$

Quantity of alcohol in 5 litres solution = 20% of 5 litres = 1 litre
Quantity of alcohol removed = 20% of 2 litres = 400 ml
Quantity of alcohol in new solution :
= (1000 - 400)ml
= 600 ml
∴ Strength of alcohol in new solution
$$\eqalign{ & = \left( {\frac{{600}}{{5000}} \times 100} \right)\% \cr & = 12\% \cr} $$

Suppose the man ordered x feet y inches of the rope. Since x and y represent inches in the miswritten and actual order respectively, so each one of x and y is less than 12.
[ $$\because $$1 feet = 12 inches]
Actual order = x feet y inches = (12x + y) inches
Miswritten order = y feet x inches = (12y + x) inches
∴ (12y + x) = 30% of (12x + y) = $$\frac{3}{10}$$ (12x + y)
⇒ 10 (12y + x) = 3 (12x + y)
⇒ 26x = 117y
⇒ $$\frac{x}{y}$$ = $$\frac{117}{26}$$ = $$\frac{9}{2}$$
Since x < 12, y < 12, so x = 9, y = 2
Hence, the man ordered 9 feet 2 inches of rope.